Chapter 3 Material Balance Applied to Oil Reservoirs 3.1 Introduction

-The Schilthuis material balance equation - Basic tools of reservoir engineering => Interpreting and predicting reservoir performance.

-Material balance

1. zero dimension this chapter 2. multi-dimension (multi-phase) reservoir simulation

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3.2 General form of the material balance equation for a hydrocarbon reservoirUnderground withdrawal (RB) = Expansion of oil and original dissolved gas (RB)(A) + Expansion of gascap gas (RB) (B) + Reduction in HCPV due to connate water expansion and decrease in the pore volume (RB) .(C)

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Expansion of oil & originally dissolved gas

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Expansion of the gascap gasExpansion of the gascap gas =gascap gas (at p) gascap (at pi)

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Change in the HCPV due to the connate water expansion & pore volume reduction

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Underground withdrawal

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The general expression for the material balance as

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where

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No initial gascap, negligible water influx

With water influx eq(3.12) becomes

Eq.(3.12) having a combination drive-all possible sources of energy.

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3.4 Reservoir Drive Mechanisms

Reservoir drive mechanism

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3.5 Solution gas drive(a) above the B.P. pressure (b) below the B.P. pressure

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Above the B.P. pressure- no initial gascap, m=0- no water flux, We=0 ; no water production, Wp=0- Rs=Rsi=Rp from eq.(3.7)

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Exercise3.1 Solution gas drive, undersaturated oil reservoir Determine R.F. Solution: FromTable2.4(p.65)

Eq(3.18)

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Table 2.4 Field PVTP(psia) Bo (Rb/STB) Rs(SCF/STB) Bg( Rb/SCF)

4000 (pi) 1.2417 5103500 1.2480 5103300 (pb) 1.2511 510 0.000873000 1.2222 450 0.00096 1.2022 401 0.00107 1.1822 352 0.00119 1.1633 304 0.001371800 1.1450 257 0.00161 1.1287 214 0.001961200 1.1115 167 0.00249 1.0940 122 0.00339 1.0763 78 0.00519300 1.0583 35 0.01066

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Bo as Function of Pressure

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Rs as Function of Pressure

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Bg and E as Function of Pressure

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Producing Gas-oil Ratio (R) as Function of Pressure

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Below B.P. pressure (Saturation oil)

Pgas liberated from saturated oil

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Exercise3.2 Solution gas drive; below bubble point pressure Reservoir-described in exercise 3.1 Pabandon = 900psia(1) R.F = f(Rp)? Conclusion?(2) Sg(free gas) = F(Pabandon)?

Solution:

(1) From eq(3.7)

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Eq(3.7) becomes

Conclusion:

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(2) the overall gas balance

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